Guru Nanak Dev Engineering College

Ludhiana

Department Of Civil Engineering

Topic –Design philosophies

Submitted By –Aryan Gupta

C.R.N - 2214006

Submitted To –Dr H.S Rai Sir

DESIGN PHILOSOPHIES

Objectives of RCC Design

To achieve a structure that will result in a Safe and Economical solution. It

consists of the following steps:

1Idealization of Structure for Analysis [The process of replacing an actual

structure with a simple system conducive to analysis is called structural

idealization].

2Estimation of Loads

3 Analysis of Idealized Structure to determine

(a) Axial Thrust

(b) Shear

(d) Deflections

4 Design of Structural Elements

5 Material Specifications and Detailing of Reinforcement

6 Detailed Structural Drawing

Design Philosophies

1 Working Stress Method

2 Ultimate Load method

3 Limit State Design

BEAM

Idealized Beam

WORKING STRESS METHOD

Fig 1: Stress Strain Curves for Concrete and Steel

Working Stress Method is an Elastic Method i.e. the material behaves Elastically

and Hook's Law is valid i.e. Stress is Proportional to Strain.

This means Strength of only that part of the material is considered which is under

Elastic Limit i.e. up to point 'A'.

The remaining Strength Beyond point A is not taken into consideration for a

particular Load.

This can be summarized as

1 Concrete is Elastic

2 Steel and Concrete act together Elastically

3 The relationship between Stress and Strain is Linear right upto failure/collapse

of the structure.

4 The Permissible Stresses for Concrete and Steel are not exceeded anywhere in

the Structure when subjected to the worst combinations of Working Loads.

5 The Sections are Designed as per the Elastic Theory of Bending assuming that

both materials i.e. concrete and steel obey the Hook's law.

6 The Elastic Theory assumes a Linear Variation of Strain and Stress i.e. Zero at

the Neutral Axis and Maximum at the extreme fibre.

Fig 2 Stress & Strain Variation Along the Depth of

Beam

Beam Section

Strain Variation Stress Variation

Beam Section Strain Variation Stress Variation

ASSUMPTIONS

1A Section which is Plane before Bending remains Plain after Bending

2 Bond between Steel and Concrete is perfect within the Elastic Limit of Steel

3 The Stress Strain Relationship of Steel and Concrete under Working Loads is

a Straight Line

4 Concrete is Elastic i.e. Stress varies Linearly from Zero at the Neutral Axis to

a maximum at the extreme fibre.

5 The Modular Ratio 'm' has the value

Where:-

σcbc :-Permissible Compressive Strength due to Bending in MPa

Permissible Stresses

Although the Stresses are kept within the Elastic Limits and most of the

Strength is not taken into account, the Stresses in Concrete and Steel are further

reduced by a Factor of Safety to account for uncertainties in the Estimation of

the Working Loads and variations in the Properties of Materials

Factor of Safety

For Concrete 3

Steel 1.78

The Permissible Stress is obtained by dividing Grade of concrete by 3 and Yield

Strength of Steel be 1.78.

Different values of Permissible Stresses for Concrete and Steel are given in

Tables 21 and 22 IS 456 - 2000 and for Shear Stresses in Table 23 of IS 456 -

2000.

Hence for properly designed Structural Elements the Stresses computed under

the action of working loads will be well within the Elastic Range. e.g.

For M20 Concrete :

Permissible Compressive Stress Permissible Stress in Concrete

in Bending Compression

For Fe250 Steel :

Permissible Tensile Stress Permissible Stress in Concrete

in Bending Tension

Structural Element is Designed for :

(1) Strength and (2) Stiffness under working loads

Strength Consists of:

(a) Compression Bending

(b) Shear

(d) Combination of above factors

Stiffness Consists of:

(a) Deflections

(b) Cracking

(d) Vibrations

Types of Loads

(a) Dead Loads

(b) Live Loads or Imposed Loads

(d) Earthquake Load

Static Loads

Dynamic Loads

DRAWBACKS

1 Concrete is not an Elastic Material. The inelastic material behaviour of

concrete actually starts from very low stresses. The Actual stress distribution

in concrete section is in fact Non Linear and cannot be described by a

Triangular Stress Diagram.

2 The Section Designed is Uneconomical as the Strength of the material

beyond the Elastic Zone is not taken into consideration.

3 Since Factor of Safety is on the Stresses under Working Loads; its not

possible to account for uncertainties associated with different types of loads

LIMIT STATE DESIGN METHOD

The object of Design based on Limit State Concept is to:

“Achieve an acceptable Probability that a Structure will not become

Unserviceable in its Life Time for the use for which it is intended, i.e. it will

not Reach to a Limit State”.

A Structure with appropriate degrees of Reliability should be able to

withstand safely all the loads that are liable to act on it throughout its life and

it should also satisfy the Serviceability Requirements, such as,

Deflections and

Cracking

It should also be able to maintain the required Structural Integrity during

and after accidents, such as:

•Fires

•Explosions

•Local Failures

In other words, all relevant Limit States must be considered in the Design to

ensure an adequate degree of “Safety and Serviceability”

Various Limit States, which needs to be examined are:

1Limit State of Collapse

The Limit State that concerns with the safety of people and/or safety of

structure are classified as “Limit State of Collapse” or Ultimate Limit State.

Following Limit States shall be verified where they are relevant

Loss of Equilibrium of the Structure or any part of it considered as a

Rigid Body

Failure due to Excessive Deformations, Rupture and Loss of Stability of

Structure.

Failure caused by Fatigue or other Time Dependent effects.

These above Limit States Corresponds to Maximum Load carrying capacity

Violation of Collapse Limit State implies Failure in the sense that a clearly

defined Limit State of Structural Usefulness has been exceeded.

However it does not mean a complete collapse

The Collapse Limit State may correspond to

•Flexure

•Compression

•Shear

•Torsion

2 Limit State of Serviceability

The Limit State that Concerns with the :

Functioning of the Structure or Structural Members under Normal use

Comfort of People orAppearance of the Construction work

Shall be Classified as Serviceability Limit State

The verification of Serviceability Limit State is based on the following aspects

(a) Deformations that affects

•Appearance

•Comfort of Users

•The Function of the Users

(b) Vibrations

•That Causes Discomfort

•That Limits the Functional Effectiveness of Structure

•Appearance

•Durability

•Functioning

Hence effectively it is a S3problem i.e. for structure to be safe, it should have

1. Strength

2. Stiffness/Serviceability

3. Stability against

Overturning

Sliding

Shrinkage

And off course

4 Economy

The Limit State approach of Design basically considers the strength of material beyond

Elastic Limit where the Stress is no longer Proportional to Stress.

It follows a Parabolic variation.

This is due to the Non-Linear behaviour of concrete as it is a Heterogeneous Material.

Figure 3 Stress Strain Curve in Limit State Design

The Limit State of Design of RCC Structures takes into account the

probabilistic and Structural Variations in the

Material properties,

Loads and

Safety Factors.

Hence the Resistance to Bending ,Shear,Torsion and Axial Loads at every

section shall not be less than the Appropriate value at that section produced

by the probable most unfavourable combination of Loads on the Structure

using the appropriate Partial Safety Factors.

The Limit State of Collapse can be expressed by the following expression:

Where

R:- Resistance or capacity of structure

L:- Loads acting on the Structure

:- Load Factor >1

: Safety Factor associated with the uncertainties in the material

properties

The Factors and are called partial safety factors.

Characteristic & Design Values and Partial Safety Factors

In order to keep the Structure Safe, Partial Safety Factors are applied to the

calculations of Loads and Material Strength.

Characteristic Strength of Material

It means that value of Strength of Material below which not more than 5% of

test results are expected to fall.

The Characteristic Strength should be in accordance with Table 2 of IS 456 –

2000

Design Value

The Characteristic Strength is determined in a laboratory by testing 15 cm

cubes in compression under controlled conditions.

But at site the conditions are not controlled.

The strength is a non-quantified quantity and due to the Errors,Mistakes and

Ignorance of workers at site, it is possible that the required strength may not be

achieved.

Hence, in order that the structure does not fail due to lack of strength of

material the Design Strength is taken into consideration.

Design Strength =

Where, fck :- Characteristic Strength of Material

m:- Partial Safety Factor appropriate to Material

The Partial Safety Factor for Material Strength is a DIVIDING Factor i.e. the

Material Strength is reduced and only corresponding loads are allowed on the

structure.

Partial Safety Factor m[as per IS 456 2000 clause 36.4.2.1] for

Concrete = 1.5

Steel = 1. 15

Characteristic Loads [Clause 36.2 of IS 456 –2000]

The term Characteristic Load means that value of Load which has a 95% Probability

of not being exceeded during the life of the structure.

For the purpose, of determining the loads the loads are taken as per ::

Dead Loads as given in IS 875 [Part 1]

Imposed Loads as given in IS 875 [Part 2]

Wind Loads as given in IS 875 [Part 3]

Snow Loads as given in IS 875 [Part 4]

Seismic Loads as given in IS 1893

Shall be assumed as Characteristic Loads

Design Values

In order that the Loads coming on the Structure never surpasses or are more

than the Characteristic Loads, the value of the load given in the code is

increased by multiplying it by a factor known as Partial Safety Factor and is

termed as “Designed Load”

Design Load Fdis given by:

Where F:- Characteristic Load

f:- Partial Safety Factor appropriate to nature of loading

•The Values of Partial Safety Factor for Load shall be taken as given in Table 18

of IS 456 –2000 i.e.

Load

Combination

Dead Load

(DL)

Imposed Load

(IL)

Wind Load

(WL)

DL + IL 1.5 1.5 -

DL + WL 1.5 - 1.5

*DL + WL 0.9 - 1.5

DL +IL + WL

1.2 1.2 1.2

* When Stability against overturning is critical or Stress Reversal is

Critical

Stress –Strain Relationship for Concrete

Fig 4 Idealized Stress Strain Curve and Stress Block Parameters for

Concrete

The Mechanical Properties of Concrete and therefore, the Stress Strain

Relationship is influenced by a number of factors such as:

Type of Aggregate and Concrete Mix

Strength of Concrete

Age of Concrete

Shape and Size of Concrete Specimen

Creep

The experimental Stress-Strain curve for concrete is too complicated to be

used in design.

The Code IS 456 –2000 has idealized it as shown in Fig 4 [Given at page 69,

Clause 38, Fig 21].

Since the concrete is heterogeneous, the curve is a Parabola i.e. Stress is not

Proportional to Strain.

As per the Code the Curve remain parabolic upto a strain of 0.002, beyond

which the strain remains constant with increasing load until a strain of 0.0035

(0.35%) has reached when the concrete is said to have failed.

Stress –Strain Relationship for Steel

Fig 5 Idealized Stress Strain Curves for Steel

Actual Stress Strain

curve for Mild Steel Idealized Stress Strain curve for

Steel having definite Yield Point

Actual Stress Strain

curve for HYSD Bars Idealized Stress Strain curve for

HYSD Bars

Fig 6 Typical Stress Strain Curves for Mild Steel and HYSD Bars

SINGLY REINFORCED SECTIONS

ATension

Compression

Clear Span

A Reinforced Concrete Flexure Member should be able to resist to the following

stresses induced due to Imposed Loads :

Tensile Stress

Compressive Stress

Shear Stress

Concrete :-

Fairly Strong in Compression

Weak in Tension

Tensile Strength taken as Zero

Steel:-

Very Strong in Tension

Steel takes up Tension in the Tensile Zone of the Flexural Member.

While Designing a Reinforced Concrete Section, the Loading,Span,Grade of

Concrete, Grade of Steel and Width of Section are usually known in advance.

The Section Dimensions and Area of Steel Bars [Reinforcing Steel] are to be

determined.

There can be no unique section for a given set of forces. There are many

possible combinations.

Thus the cost will decide the final design

ASSUMPTIONS

Design for the Limit State of Collapse in Flexure shall be based on the assumptions

as per IS 456 –2000; Clause 38.1, p 69

Plane Sections Normal to the Axis remain Plane after Bending

The Maximum Strain in Concrete at the outermost Compression Fibre is taken as

0.0035 in Bending

The Relationship between Stress-Strain distribution in Concrete is Parabolic upto

a Strain of 0.002 and then constant upto a Strain of 0.0035 at which the concrete is

said to have failed [IS 456 –2000, Fig 21, pp. 69]

For Design purpose the Compressive Strength of Concrete is taken as 0.67 times

the Characteristic Strength of Concrete. A Partial Safety Factor m= 1.5 shall be

applied in addition to this.

The Tensile Strength of Concrete is ignored

The Stress in the Reinforcement is derived from the representative Stress-Strain

Curve for the type of Steel used. The typical Curves are given in Fig 23 of IS 456 –

2000;pp.70

For Design Purposes the Partial Safety Factor of m= 1.15 shall be applied to the

Characteristic Strength of Steel

The Maximum Strain in Tension Reinforcement in the Section at Failure should

not be less than the following i.e.

Bending of Beams

We know that

fcr = 0.7fCK

MOMENT OF RESISTANCE

Neutral Axis

Fig 8: Stress Block Parameters

Consider a Simply Supported Beam subjected to Bending under factored loads.

For Equilibrium total force of Compression must be equal to the total force of

Tension i.e. C = T

The applied Bending Moment at Collapse i.e. Factored Bending Moment is equal to

the Resisting Moment of the Section provided by the Internal Stresses.

This called the Ultimate Moment of Resistance.

Now Force = C = C1+ C2

C = Force x Area

And MR= Force x Lever Arm

The portion above the Neutral Axis is in Compression and the Strain is

proportional to distance from Neutral Axis (NA) to the Extreme Compression

Fibre i.e. Zero at the NA to a Maximum at the Extreme fibre.

The cross section below the NA is in Tension and hence the Concrete is

assumed to have Cracked.

All the Tensile Stresses are supposed to be borne by steel bars and stresses in

all the steel bars are equal.

The resultant Tensile Force thus acts at the Centroid of the Reinforcing Bars.

The distance from the Extreme Compression Fibre to the centroid of the

Reinforcing Bars i.e. line of action of Tensile Force is called the Effective Depth

'd'.

Now,

Maximum Compressive Stress in Concrete without Safety Factor

= 0.67 fCK [Assumption 4]

Let,

X1: Depth of Parabolic Portion

X2: Depth of Rectangular Portion

From,

Similar Triangles of Strain Diagram,

Depth of Parabolic Portion is

Or,

Depth of Rectangular Portion

X2=XU- X1

Force of Compression

•

Parabolic Block :-

C1= Stress x Area

= (0.67 fCK ) ×(2/3 X1. b)

= (0.67 fCK ) ×(2/3 . XU. b)

C1= 0.255 fCK .b . XU

•

Rectangular Block:-

C2= Stress x Area

= (0.67 fCK ) ×( X2. b)

= (0.67 fCK ) ×( XU. b)

C2= 0.287 fCK .b . XU

•

Hence Total Force of Compression without Partial Safety Factor

CO= C1 + C2

= 0.255 fCK .b . XU+ 0.287 fCK .b . XU

CO= 0.542 fCK b XU{Without Partial Safety Factor}

Now Applying Partial Safety Factor of 1.5 the Design Force of Compression is:-

C = 0.36 fCK b XU

Now,

Moment of Resistance = Force ×Lever Arm

Lever Arm = Z = d ˗ a

Where,

'a' is the distance of line of action of force of compression from the extreme top

fibre.

To determine 'a' take moment of all the forces about top extreme fibre. i.e.

CO ×a = 0.225 fCK .b . XU2

a = 0.42 XU

Where,

XU: Depth of Neutral Axis from Top Fibre

B : Width of the Section

DEPTH OF NEUTRALAXIS

Depth of NA can be obtained by considering the equilibrium of normal force i.e.

Force of Compression = Force of Tension

Resultant Force of Compression

C = Average Stress ×Area

C = 0.36 fCK b XU

Resultant Force of Tension

T = 0.87 fYAt

Now

C = T

0.36 fCK b XU = 0.87 fYAt

OR,

Where,

At= Area of Tension Steel

LEVER ARM

The forces of Compression and Tension forms a Couple.

The distance between the lines of action of these two forces is called the Lever

Arm and is denoted by 'Z'.

The equation of equilibrium ΣM = 0 is satisfied by equating the factored

Bending Moment to the Moment of Resistance offered by either Force of

Compression or Force of Tension.

Lever Arm Z = d ˗a

Z = d ˗0.42 XU

Now,

Moment of Resistance

Now,

Moment of Resistance w. r. t. Concrete

MRC = Compressive Force ×Lever Arm

MRC = 0.36 fCK b XU. Z

Moment of Resistance w. r. t. Steel

MRt = Compressive Force ×Lever Arm

MRt = 0.87 fYAt. Z

MODES OF FAILURE

Balanced Section :-

If the ratio of Steel to Concrete in a beam is such that the maximum strain in

concrete and steel reach simultaneously, a sudden failure would occur with less

alarming deflections.

Such a beam is referred to as a Balanced Reinforced Beam.

Under Reinforced Beam :-

When the amount of steel is kept less than that in the Balanced Section, the NA

moves upwards so as to reduce the area under compression to maintain the

Equilibrium Condition i.e.

Force of Compression is equal to the Force of Tension. [This is because the Force

of tension becomes less than the Force of Compression and hence the Force of

Compression has to be reduced]

In this process the Centre of Gravity of compressive forces also shifts upwards.

Under increasing Bending Moments Steel is strained beyond Yield Point and the

Maximum Strain in concrete remains less than 0.35% i.e. 0.0035.

If the beam is further loaded, the strain the section increases. Once the steel has

yielded it does not take any additional stresses for the additional strain and the

total force of tension remains constant.

However compressive stresses in concrete increases with the additional strain.

Thus the NA and Centre of Gravity of Compressive Forces further shifts

upwards to maintain Equilibrium.

This results in an increase in the Moment of Resistance of the Beam. This

process of shift in the NA continues until maximum strain reaches its Ultimate

Value i.e. 0.35% and the Concrete Crushes.

Such a beam is referred to as "Under Reinforced Beam".

The Failure is called Tension Failure because Yielding of Steel was responsible

for higher strains in concrete resulting in its failure.

Over Reinforced Beam :-

When the amount of steel is kept more than that in the Balanced, the NA tends

to move downwards and the Strain in Steel remains in Elastic Region.

If the beam is further loaded the stresses in steel keeps on increasing and so the

force of tension.

Here the force of tension is more than that of compression, and hence to

maintain the equilibrium of tensile and compressive forces the area of concrete

resisting compression has to increase so as to increase the force of compression.

In this process the NA furher shifts downwards until maximum strain in

concrete reaches its ultimate value of 0.35% and concrete crushes. The Steel is

well within Elastic Limits.

Such a beam is referred to as an "Over Reinforced Beam" and the failure as

Compression Failure

MAXIMUM DEPTH OF NEUTRAL AXIS

A compression failure in a Over Reinforced Beam is a Brittle Failure.

The Maximum Depth of NA is therefore limited to ensure that the Steel will

reach its Yield Point before Concrete fails in Compression, so that a brittle

Failure is avoided.

Let the Limiting Value of the depth of NA be XU Lim.

When,

XU= XU Lim [Balanced Section ]

XU< XU Lim [Under Reinforced Section ]

XU> XU Lim [Over Reinforced Section ]

The Limiting Value of Depth of NA XU Lim.for different grades of steel can be

obtained from Strain Diagram as shown in Fig 8.

From Similar Triangles

Where

E = 2 ×105N/mm2

The Limiting values of Depth of NA for different grades of steel are given in Table 1

From Fig 8

0.0035

0.002

XU Lim

Grade of Steel fY

(N/mm2)

XU Lim

250 0.53 d

415 0.48 d

500 0.46 d

550 0.44 d

Table 1 Maximum Depth of Neutral Axis

LIMITING VALUES OF TENSION STEELAND MOMENT OF RESISTANCE

Since the maximum depth of NA is limited the maximum value of moment of resistance

is also limited i.e.

MU Lim w. r. t Concrete = 0.36 fCK b XU. Z

MU Lim = 0.36 fCK b XU Lim (d ˗0.42 XU Lim ) {Balanced Section}

MU Lim w. r. t Steel = 0.87 fYAt. Z

MU Lim = 0.87 fYAt(d ˗0.42 XU Lim ) { Balanced Section}

For a given Rectangular Beam Section, the Limiting Values of MU Lim depends on

the Grade of Concrete and the Grade of Steel.

The Values of Limiting Moment of Resistance with respect to different Grades of

Concrete and Steel are given in Table 2.

Grade of

Concrete Fe 250 Fe 415 Fe 500 Fe 550

General 0.148fCK b d20.138fCK b d20.133fCK b d20.130fCK b d2

M 20 2.96 b d22.76 b d22.66 b d22.60 b d2

M 25 3.70 b d23.45 b d23.33 b d23.25 b d2

M 30 4.4 b d24.14 b d23.99 b d23.90 b d2

Table 2: Limiting Moment of Resistance (N-mm)

The percentage of Tensile Reinforcement Corresponding to Limiting Moment of

Resistance is obtained by equating the forces of tension and compression

0.87 fYAt= 0.36 fCK b XU Lim

Let,

The limiting values of tensile reinforcement percentage corresponding to different

grades of steel and concrete in a Singly Reinforced Beam are given in Table 3.

Grade of

Concrete

Percentage of Tensile Steel

fCK (N/mm2)

Fe 250 N/mm

2Fe 415 N/mm2Fe 500 N/mm2

Fe 550 N/mm2

M 20 1.76 0.96 0.76 0.66

M 25 2.20 1.19 0.94 0.83

M 30 2.64 1.43 1.13 1.00

Table 3: Limiting Tensile Steel in Rectangular Section (%)

Minimum and Maximum Tension Reinforcement

Minimum Reinforcement :[ As per Clause 26.5.1.1; pp. 46; of IS 456 - 2000 ]

The minimum area of tension reinforcement should not be less than that given

by the following :

Where,

AS= Minimum Area of Tension Reinforcement

b = Breadth of Beam or breadth of Web of T-Beam

d = Effective depth of Beam

fY= Characteristic Strength of Steel Reinforcement in N/mm2

Maximum Reinforcement :

•The maximum area of tension reinforcement should not exceed 4% of the

Gross Cross-Sectional area of beam to avoid difficulty in placing and

compacting concrete properly in the formwork i.e.

ASM > 0.04 b D

THANK

YOU

For given your Time